Spring 2004 Math 253/501–503
13 Multiple Integrals
13.10 Triple Integrals in Cylindrical
and Spherical Coordinates
c
Thu, 11/Mar
2004,
Art Belmonte
Solution
The θ - and φ-specifications indicate that the solid lies inside a
cone. For fixed θ and φ, the innermost variable ρ varies from
ρ = 0 (the origin) to ρ = sec φ or z = ρ cos φ = 1, a horizontal
plane. What we have here is an HCFT: a half-cone with a flat top!
The volume is
ZZZ
Z 2π Z π/3 Z sec φ
V =
1 dV =
ρ 2 sin φ dρ dφ dθ π ≈ 3.14 cm3 .
Summary
Recall the relationships between rectangular (x, y, z), cylindrical
(r, θ, z), and spherical (ρ, θ, φ) coordinates.
1
0.8
r 2 = x 2 + y2
tan θ =
y
x
0.6
z=z
0.4
0.2
Note that (typically) r ≥ 0 and 0 ≤ θ ≤ 2π.
0
• Spherical / rectangular relationships
x = ρ sin φ cos θ
r = ρ sin φ
0
z
y = r sin θ
0
Stewart 853/4: HCFT
(half−cone, flat top)
• Cylindrical / rectangular relationships
x = r cos θ
0
E
−1
y = ρ sin φ sin θ
0
x
y
z = ρ cos φ
−1
0
1
1
%
% Stewart 853/4
%
syms p r t
volume = int(int(int(1 * rˆ2*sin(p), ...
r,0,sec(p)), p,0,pi/3), t,0,2*pi);
pretty(volume)
ρ2 = x 2 + y2 + z2
Note that ρ ≥ 0, 0 ≤ φ ≤ π, and 0 ≤ θ ≤ 2π.
• Here are three specifications of the volume differential dV :
rectangular, cylindrical, and spherical, respectively.
pi
d x d y dz
=
r dz dr dθ
=
ρ 2 sin φ dρ dφ dθ
floated = eval(volume)
floated =
3.1416
%
When we cover Jacobian matrices and determinants in
Section 13.11, we’ll see how the extra factors r and ρ 2 sin φ
are obtained. Until then, just look at the figures 3 and 7 in
Section 13.10 of Stewart for a geometrical explanation.
echo off; diary off
853/6
ZZZ q
Evaluate
Hand / MATLAB Examples
x 2 + y 2 dV , where E is the solid bounded by
E
the paraboloid z = 9 − x 2 − y 2 and the x y-plane.
Aids to our hand work are the TAMUCALC Muint menu and the
smi command. The former contains a variety of multiple integral
templates that facilitate typing. The latter shows the steps involved Solution
in integration.
Here is a picture of the solid.
Pay careful attention to the 2-D and 3-D graphics in the examples.
Stewart 853/6
These will help you to set up the limits of integration in your triple
integrals. With practice, you’ll be able to draw rough sketches that
will suffice. In upcoming computer assignments, you will draw
8
some graphs yourselves with MATLAB.
z
6
4
853/4
2
Sketch the solid whose volume is given by the integral
Z 2π Z π/3 Z sec φ
ρ 2 sin φ dρ dφ dθ , then compute this volume.
0
0
0
2
2
0
y
0
1
0
−2
−2
x
854/18
Switch to cylindrical coordinates. The paraboloid intersects the
x y-plane in the circle r 2 = x 2 + y 2 = 9, z = 0. Hence the value
of the integral is
Z
2π
0
Z
3 Z 9−r 2
0
ZZZ
Evaluate
xe
x 2 +y 2 +z 2
2
dV , where E is the solid that lies
E
r · r dz dr dθ =
0
324
5 π
between the spheres x 2 + y 2 + z 2 = 1 and x 2 + y 2 + z 2 = 4 in
the first octant.
≈ 203.575.
%
% Stewart 853/6
%
syms r t z
volume = int(int(int(r * r, ...
z,0,9-rˆ2), r,0,3), t,0,2*pi);
pretty(volume)
Solution
Here is a picture, with a transparent outer sphere through which
we see the inner sphere.
324/5 pi
floated = eval(volume)
floated =
203.5752
%
Stewart 854/18
2
echo off; diary off
1.5
z
854/13
1
0.5
Find the mass and center of mass of the solid E bounded by the
paraboloid z = 4x 2 + 4y 2 and the plane z = a (a > 0) if E has
constant density δ = k.
0
1
0
1
2
2
x
y
Switch to spherical coordinates, then evaluate the integral.
Solution
Z
When the paraboloid
intersects the plane, we have a = z = 4r 2 ,
1√
whence r = 2 a. Here are the mass m and center of mass CM.
ZZZ
m =
δ dV
Z
=
=
CM
E
2π
√
Z
[x̄, ȳ, z̄]
ZZZ
=
1
m
=
4r 2
0
0
1 ka 2 π
8
=
=
a/2 Z a
0
=
π/2 Z π/2 Z 2
0
ρ sin φ cos θ · eρ · ρ 2 sin φ dρ dφ dθ
1
π 16
e − e ≈ 1.745 × 106
16
%
% Stewart 854/18
%
syms p r t
value = int(int(int(r*sin(p)*cos(t) * exp(rˆ4) * rˆ2*sin(p), ...
r,1,2), p,0,pi/2), t,0,pi/2);
pretty(factor(value))
k · r dz dr dθ
[x, y, z] δ dV
E
1
Z
1 ka 2 π
0
h8
i
2
0, 0, 3 a
2π
1/16 pi (exp(16) - exp(1))
√
a/2 Z a
Z
0
4
4r 2
floated = eval(value)
floated =
1.7448e+06
%
[x, y, z] · k · r dz dr dθ
echo off; diary off
854/21
Stewart 854/13
Stewart 854/13: Center of mass
a
Find the volume of the solid that lies above the cone φ =
below the offset sphere ρ = 4 cos φ.
a
z
CM = ( 0, 0, 2a / 3 )
Solution
0
y
a1/2 / 2
0
1/2
a
0
0
/2
x
0
Here is a picture of the solid.
0
2
π
3
and
Switching to cylindrical coordinates we have
Stewart 854/21: Offset sphere and half−cone
3
z
π/2 Z 1 Z r
Z
4
0
2
−1
0
1
−1 0
y
1
x
0
r cos θ · r sin θ · z · r dz dr dθ =
1
≈ 0.0104.
96
%
% Stewart 854/34
%
syms r t z
value = int(int(int( r*cos(t) * r*sin(t) * z
z,rˆ2,r), r,0,1), t,0,pi/2);
pretty(value)
1
0
r2
*
r, ...
1/96
The volume is
V
floated = eval(value)
floated =
0.0104
%
ZZZ
=
1 dV
E
Z
Z
2π
=
0
=
π/3 Z 4 cos φ
0
echo off; diary off
1 · ρ 2 sin φ dρ dφ dθ
0
854/36
10π ≈ 31.4 cm3 .
Evaluate the integral
Z 3 Z √9−y 2 Z √18−x 2 −y 2
%
% Stewart 854/21
%
syms p r t
volume = int(int(int(1 * rˆ2*sin(p), ...
r,0,4*cos(p)), p,0,pi/3), t,0,2*pi);
pretty(volume)
0
√
0
x 2 +y 2
x 2 + y 2 + z 2 dz d x d y
by changing to spherical coordinates.
10 pi
floated = eval(volume)
floated =
31.4159
%
Solution
The p
region of integration lies in the first octant, with the cone
z = x 2 + y 2 below and the sphere x 2 + y 2 + z 2 = 18 above.
Here is a picture showing boundary surfaces of the solid.
echo off; diary off
854/34
√
1Z
Z
1−y 2
Z √x 2 +y 2
Evaluate the integral
0
0
4
x yz dz d x d y by
Sphere
xz−plane
3
z
changing to cylindrical coordinates.
x 2 +y 2
Stewart 854/36
Solution
2
1
Cone
0
0
The region of integration lies in the first octant, with
p the
paraboloid z = x 2 + y 2 below and the cone z = x 2 + y 2 above.
Here is a picture showing boundary surfaces of the solid.
1
2
0
x
2
1
y
p
The cone is ρ cos φ = z = x 2 + y 2 = r = ρ sin φ, whence
tan φ = 1 or φ = π4 . Switching to spherical coordinates we have
Stewart 854/34
Z
1
Cone
√
18
π/2 Z π/4 Z
0.8
0
0
0
486π √
2 − 1 ≈ 126.485.
5
z
0.6
ρ 2 ·ρ 2 sin φ dρ dφ dθ =
Paraboloid
0.4
%
% Stewart 854/36
%
syms p r t
value = int(int(int( rˆ2 * rˆ2*sin(p), ...
r,0,sqrt(18)), p,0,pi/4), t,0,pi/2);
pretty(factor(value))
0.2
0
0
xz−plane
1
0.5
0.5
1
x
0
y
3
1/2
486/5 pi (2
- 1)
floated = eval(value)
floated =
126.4854
%
echo off; diary off
854/38
Show that
Z ∞Z
∞
Z
∞
−∞ −∞ −∞
q
x 2 + y 2 + z 2 e−x
2 −y 2 −z 2
d x d y dz = 2π.
Solution
The region of integration is all of 3-D space. So integrate over a
spherical ball of radius a centered at the origin, then take a limit of
the result as a → ∞.
Z 2π Z π Z a
2
lim
ρe−ρ ρ 2 sin φ dρ dφ dθ
a→∞ 0
02 0
2π ea − a 2 − 1
= lim
2
a→∞
ea
= 2π
%
% Stewart 854/38
%
syms a positive
syms p r t
sphere a value = int(int(int( r*exp(-rˆ2) * rˆ2*sin(p), ...
r,0,a), p,0,pi), t,0,2*pi);
pretty(sphere a value)
2
2
(a˜ + 1 - exp(a˜ )) pi
-2 ----------------------2
exp(a˜ )
limiting value = limit(sphere a value, a, inf);
pretty(limiting value)
2 pi
%
echo off; diary off
4